ΔH for the reaction N₂ + 3H₂ ⇋ 2NH₃

The correct option is ΔE – 2RT.

To understand why this is the correct relation, we start with the thermodynamic relationship between enthalpy change ($\mathrm{\Delta H}$) and internal energy change ($\mathrm{\Delta E}$) for a chemical reaction involving ideal gases:

$\mathrm{\Delta H}=\mathrm{\Delta E}+\Delta n_{g}RT$

Here:
• $\mathrm{\Delta H}$ is the change in enthalpy.
• $\mathrm{\Delta E}$ is the change in internal energy.
• $\Delta n_g$ is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature.

Let us analyze the given balanced chemical reaction:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

First, we find the number of moles of gaseous products:
$n_{\text{products}}=2$ (from $2N{H}_3$)

Next, we find the number of moles of gaseous reactants:
$n_{\text{reactants}}=1+3=4$ (from $1N_2$ and $3H_2$)

Now, we calculate $\Delta n_g$:
$\Delta n_g=n_{\text{products}}-n_{\text{reactants}}$
$\Delta n_g=2-4=-2$

Substituting the value of $\Delta n_g=-2$ into our original equation:
$\mathrm{\Delta H}=\mathrm{\Delta E}+(-2)RT$
$\mathrm{\Delta H}=\mathrm{\Delta E}-2RT$

Therefore, the relationship between enthalpy and internal energy for this reaction is represented by the expression ΔE – 2RT.